An Elementary Study of Chemistry, William McPherson [icecream ebook reader txt] 📗
- Author: William McPherson
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Method of calculation. The densities of the various gases in the first column of this table are determined by experiment, and are fairly accurate but not entirely so. By multiplying these densities by 28.9 the molecular weights of the compounds as given in the second column are obtained. By chemical analysis it is possible to determine the percentage composition of these substances, and the percentages of nitrogen in them as determined by analysis are given in the third column. If each of these molecular weights is multiplied in turn by the percentage of nitrogen in the compound, the product will be the weight of the nitrogen in the molecular weight of the compound. This will be the sum of the weights of the nitrogen atoms in the molecule. These values are given in the fourth column in the table.
If a large number of compounds containing nitrogen are studied in this way, it is probable that there will be included in the list at least one substance whose molecule contains a single nitrogen atom. In this case the number in the fourth column will be the approximate atomic weight of nitrogen. On comparing the values for nitrogen in the table it will be seen that a number which is approximately 14 is the smallest, and that the others are multiples of this. These compounds of higher value, therefore, contain more than one nitrogen atom in the molecule.
Accurate determination of atomic weights. Molecular weights cannot be determined very accurately, and consequently the part in them due to nitrogen is a little uncertain, as will be seen in the table. All we can tell by this method is that the true weight is very near 14. The equivalent can however be determined very accurately, and we have seen that it is some multiple or submultiple of the true atomic weight. Since molecular-weight determinations have shown that in the case of nitrogen the atomic weight is near 14, and we have found the equivalent to be 7.02, it is evident that the true atomic weight is twice the equivalent, or 7.02 × 2 = 14.04.
Summary. These, then, are the steps necessary to establish the atomic weight of an element.
1. Determine the equivalent accurately by analysis.
2. Determine the molecular weight of a large number of compounds of the element, and by analysis the part of the molecular weight due to the element. The smallest number so obtained will be approximately the atomic weight.
3. Multiply the equivalent by the small whole number (usually 1, 2, or 3), which will make a number very close to the approximate atomic weight. The figure so obtained will be the true atomic weight.
Molecular weights of the elements. It will be noticed that the molecular weight of nitrogen obtained by multiplying its density by 28.9 is 28.08. Yet the atomic weight of nitrogen as deduced from a study of its gaseous compounds is 14.04. The simplest explanation that can be given for this is that the gaseous nitrogen is made up of molecules, each of which contains two atoms. In this respect it resembles oxygen; for we have seen that an entirely different line of reasoning leads us to believe that the molecule of oxygen contains two atoms. When we wish to indicate molecules of these gases the symbols N2 and O2 should be used. When we desire to merely show the weights taking part in a reaction this is not necessary.
The vapor densities of many of the elements show that, like oxygen and nitrogen, their molecules consist of two atoms. In other cases, particularly among the metals, the molecule and the atom are identical. Still other elements have four atoms in their molecules.
While oxygen contains two atoms in its molecules, a study of ozone has led to the conclusion that it has three. The formation of ozone from oxygen can therefore be represented by the equation
Other methods of determining molecular weights. It will be noticed that Avogadro's law gives us a method by which we can determine the relative weights of the molecules of two gases because it enables us to tell when we are dealing with an equal number of the two kinds of molecules. If by any other means we can get this information, we can make use of the knowledge so gained to determine the molecular weights of the two substances.
Raoult's laws. Two laws have been discovered which give us just such information. They are known as Raoult's laws, and can be stated as follows:
1. When weights of substances which are proportional to their molecular weights are dissolved in the same weight of solvent, the rise of the boiling point is the same in each case.
2. When weights of substances which are proportional to their molecular weights are dissolved in the same weight of solvent, the lowering of the freezing point is the same in each case.
By taking advantage of these laws it is possible to determine when two solutions contain the same number of molecules of two dissolved substances, and consequently the relative molecular weights of the two substances.
Law of Dulong and Petit. In 1819 Dulong and Petit discovered a very interesting relation between the atomic weight of an element and its specific heat, which holds true for elements in the solid state. If equal weights of two solids, say, lead and silver, are heated through the same range of temperature, as from 10° to 20°, it is found that very different amounts of heat are required. The amount of heat required to change the temperature of a solid or a liquid by a definite amount compared with the amount required to change the temperature of an equal weight of water by the same amount is called its specific heat. Dulong and Petit discovered the following law: The specific heat of an element in the solid form multiplied by its atomic weight is approximately equal to the constant 6.25. That is,
Consequently,
This law is not very accurate, but it is often possible by means of it to decide upon what multiple of the equivalent is the real atomic weight. Thus the specific heat of iron is found by experiment to be 0.112, and its equivalent is 27.95. 6.25 ÷ 0.112 = 55.8. We see, therefore, that the atomic weight is twice the equivalent, or 55.9.
How formulas are determined. It will be well in connection with molecular weights to consider how the formula of a compound is decided upon, for the two subjects are very closely associated. Some examples will make clear the method followed.
The molecular weight of a substance containing hydrogen and chlorine was 36.4. By analysis 36.4 parts of the substance was found to contain 1 part of hydrogen and 35.4 parts of chlorine. As these are the simple atomic weights of the two elements, the formula of the compound must be HCl.
A substance consisting of oxygen and hydrogen was found to have a molecular weight of 34. Analysis showed that in 34 parts of the substance there were 2 parts of hydrogen and 32 parts of oxygen. Dividing these figures by the atomic weights of the two elements, we get 2 ÷ 1 = 2 for H; 32 ÷ 16 = 2 for O. The formula is therefore H2O2.
A substance containing 2.04% H, 32.6% S, and 65.3% O was found to have a molecular weight of 98. In these 98 parts of the substance there are 98 × 2.04% = 2 parts of H, 98 × 32.6% = 32 parts of S, and 98 × 65.3% = 64 parts of O. If the molecule weighs 98, the hydrogen atoms present must together weigh 2, the sulphur atoms 32, and the oxygen atoms 64. Dividing these figures by the respective atomic weights of the three elements, we have, for H, 2 ÷ 1 = 2 atoms; for S, 32 ÷ 32 = 1 atom; for O, 64 ÷ 16 = 4 atoms. Hence the formula is H2SO4.
We have, then, this general procedure: Find the percentage composition of the substance and also its molecular weight. Multiply the molecular weight successively by the percentage of each element present, to find the amount of the element in the molecular weight of the compound. The figures so obtained will be the respective parts of the molecular weight due to the several atoms. Divide by the atomic weights of the respective elements, and the quotient will be the number of atoms present.
Avogadro's hypothesis and chemical calculations. This law simplifies many chemical calculations.
1. Application to volume relations in gaseous reactions. Since equal volumes of gases contain an equal number of molecules, it follows that when an equal number of gaseous molecules of two or more gases take part in a reaction, the reaction will involve equal volumes of the gases. In the equation
since 1 molecule of each of the gases CO2 and CO is set free from each molecule of oxalic acid, the two substances must always be set free in equal volumes.
Acetylene burns in accordance with the equation
Hence 2 volumes of acetylene will react with 5 volumes of oxygen to form 4 volumes of carbon dioxide and 2 volumes of steam. That the volume relations may be correct a gaseous element must be given its molecular formula. Thus oxygen must be written O2 and not 2O.
2. Application to weights of gases. It will be recalled that the molecular weight of a gas is determined by ascertaining the weight of 22.4 l. of the gas. This weight in grams is called the gram-molecular weight of a gas. If the molecular weight of any gas is known, the weight of a liter of the gas under standard conditions may be determined by dividing its gram-molecular weight by 22.4. Thus the gram-molecular weight of a hydrochloric acid gas is 36.458. A liter of the gas will therefore weigh 36.458 ÷ 22.4 = 1.627 g.
EXERCISES1. From the following data calculate the atomic weight of sulphur. The equivalent, as obtained by an analysis of sulphur dioxide, is 16.03. The densities and compositions of a number of compounds containing sulphur are as follows:
2. Calculate the formulas for compounds of the following compositions:
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